09. Vertical soft landing on the Moon

A lunar lander must fire its engine to touch down at zero speed. Does it save more fuel by braking gently early, or by waiting and braking hard at the end? This problem has an analytic solution for comparison.

Item	Definition
States	Altitude $h$, velocity $v$ (positive upward)
Control	Thrust $T$
Objective	Minimize the thrust integral, proportional to propellant use
Phases	Two phases, 12 nodes each; free knot and final times
New concepts	An integral objective and an optimized switching time

Situation

The lander starts 100 m above the surface, descending at 10 m/s. Lunar gravity is 1.62 m/s². Its upward engine thrust ranges from zero to 4000 N. Mass is held constant at 1000 kg, approximating a maneuver whose propellant use is small relative to vehicle mass.

Propellant use is proportional to impulse, the time integral of thrust. We split time into a coast phase and a braking phase, leaving the switching time to optimization.

Formulation

Dynamics

$$\dot h=v,\qquad\dot v=\frac Tm-g.$$

Upward is positive, so descent has $v<0$. Both phases use the same equations.

Phases and linkage

At the unknown switch $t_1$, altitude and velocity remain continuous:

$$h^{(1)}(t_1)=h^{(2)}(t_1),\qquad v^{(1)}(t_1)=v^{(2)}(t_1).$$

Thrust may jump. Both $t_1$ and $t_f$ are free.

Boundary conditions and constraints

$$h(0)=100,\quad v(0)=-10,\qquad h(t_f)=v(t_f)=0,$$ $$0\le T\le4000\ \mathrm N,\qquad0\le h\le100\ \mathrm m,\qquad v\le0.$$

The lander cannot go underground, rise above its initial altitude, or reverse into upward flight.

Objective

$$J=\frac1{T_{\max}t_u}\int_0^{t_f}T\,dt,\qquad t_u=\sqrt{h_0/g}\approx7.86\ \mathrm s.$$

The constant factor makes the objective roughly order one without changing the optimum. This is a running cost (cost forms).

Parameters and scaling

Use $h_0=100$ m, $v_0=-10$ m/s, $g=1.62$ m/s², $m=1000$ kg, and $T_{\max}=4000$ N. Representative units are 100 m, $\sqrt{h_0g}\approx12.7$ m/s, $t_u$, and 4000 N.

Initial guess

Use the analytic solution below: free fall followed by maximum thrust, including its switching and final times. This checks whether the numerical solution preserves the analytic answer.

Results

Altitude, velocity, and thrust. Blue and orange identify the two 12-node phases. Black dashed curves are analytic; the vertical dashed line marks ignition, and the red line is maximum thrust.

Free fall followed by powered braking. Flame indicates thrust; velocity and thrust histories show the current time alongside the analytic solution.

• Thrust is zero throughout phase 1 and 4000 N throughout phase 2.
• Ignition time: 3.632 s, at altitude 53.0 m and velocity -15.88 m/s.
• Touchdown time: 10.305 s; impulse: 26,695 N·s.
• Analytic ignition and touchdown times are 3.632 s and 10.305 s, matching the numerical result.

Explanation

Less fuel means less time in this model

Integrate the velocity equation and use the endpoint conditions:

$$\int_0^{t_f}T\,dt=m\bigl(|v_0|+gt_f\bigr).$$

The initial descent speed must be canceled regardless of the thrust schedule. The other term is gravity loss, which increases with time spent airborne. Thus minimizing fuel here is equivalent to minimizing touchdown time.

The fastest safe descent uses zero thrust as long as possible, then maximum upward braking. Intermediate thrust prolongs the maneuver and increases gravity loss. The optimum is bang-bang control.

Calculate the switching time

Let $b=T_{\max}/m-g=2.38$ m/s², and let $v_s$ be the positive magnitude of descent velocity at ignition. Free-fall distance plus braking distance equals initial altitude:

$$\frac{v_s^2-v_0^2}{2g}+\frac{v_s^2}{2b}=h_0.$$

Therefore

$$v_s=\sqrt{\frac{(v_0^2+2gh_0)b}{b+g}},\qquad t_1=\frac{v_s-|v_0|}{g},\qquad t_f=t_1+\frac{v_s}{b}.$$

For comparison, constant gentle braking from the start needs deceleration 0.5 m/s² and takes 20 s. Its impulse is 42 400 N·s, about 1.6 times the optimum.

Why optimize the knot?

Thrust jumps from zero to 4000 N at ignition. Putting that jump at a free knot lets each phase use a constant control without polynomial oscillations. Unlike Example 08, the switching time is determined by optimization.

Try it

• Lower max_thrust to 3000 N. Compare the earlier ignition, longer duration, and larger impulse with the formulas.
• Set initial_velocity to zero and inspect the free-fall phase.
• Replace the thrust guess [0, p["max_thrust"]] by [2000, 2000] and check whether the solver returns to the same solution.

Run

uv run python examples/09_soft_landing.py

View the Python source on GitHub.